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NEW!!! Tough and tricky exponents and roots questions
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Bunuel
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NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  12 Jan 2012, 02:50
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Exponents and roots problems are very common on the GMAT. So, it's extremely important to know how to manipulate them, how to factor out, take roots, multiply, divide, etc. Below are 11 problems to test your skills. Please post your thought process/solutions along with the answers.

I'll post OA's with detailed solutions tomorrow. Good luck.

1. If \(357^x*117^y=a\), where \(x\) and \(y\) are positive integers, what is the units digit of \(a\)?
(1) \(100<y^2<x^2<169\)
(2) \(x^2-y^2=23\)

Solution: tough-and-tricky-exponents-and-roots-questions-125967.html#p1029239

2. If x, y, and z are positive integers and \(xyz=2,700\). Is \(\sqrt{x}\) an integer?
(1) \(y\) is an even perfect square and \(z\) is an odd perfect cube.
(2) \(\sqrt{z}\) is not an integer.

Solution: tough-and-tricky-exponents-and-roots-questions-125967.html#p1029240

3. If \(x>y>0\) then what is the value of \(\frac{\sqrt{2x}+\sqrt{2y}}{x-y}\)?
(1) \(x+y=4+2\sqrt{xy}\)
(2) \(x-y=9\)

Solution: tough-and-tricky-exponents-and-roots-questions-125967.html#p1029241

4. If \(xyz\neq{0}\) is \((x^{-4})*(\sqrt[3]{y})*(z^{-2})<0\)?
(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\)
(2) \(y^3>\frac{1}{z{^4}}\)

Solution: tough-and-tricky-exponents-and-roots-questions-125967.html#p1029242

5. If \(x\) and \(y\) are negative integers, then what is the value of \(xy\)?
(1) \(x^y=\frac{1}{81}\)
(2) \(y^x=-\frac{1}{64}\)

Solution: tough-and-tricky-exponents-and-roots-questions-125967.html#p1029243

6. If \(x>{0}\) then what is the value of \(y^x\)?
(1) \(\frac{4^{(x+y)^2}}{4^{(x-y)^2}}=128^{xy}\)
(2) \(x\neq{1}\) and \(x^y=1\)

Solution: tough-and-tricky-exponents-and-roots-questions-125967.html#p1029244

7. If \(x\) is a positive integer is \(\sqrt{x}\) an integer?
(1) \(\sqrt{7*x}\) is an integer
(2) \(\sqrt{9*x}\) is not an integer

Solution: tough-and-tricky-exponents-and-roots-questions-125967-20.html#p1029245

8. What is the value of \(x^2+y^3\)?
(1) \(x^6+y^9=0\)
(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\)

Solution: tough-and-tricky-exponents-and-roots-questions-125967-20.html#p1029246

9. If \(x\), \(y\) and \(z\) are non-zero numbers, what is the value of \(\frac{x^3+y^3+z^3}{xyz}\)?
(1) \(xyz=-6\)
(2) \(x+y+z=0\)

Solution: tough-and-tricky-exponents-and-roots-questions-125967-20.html#p1029247

10. If \(x\) and \(y\) are non-negative integers and \(x+y>0\) is \((x+y)^{xy}\) an even integer?
(1) \(2^{x-y}=\sqrt[(x+y)]{16}\)
(2) \(2^x+3^y=\sqrt[(x+y)]{25}\)

Solution: tough-and-tricky-exponents-and-roots-questions-125967-20.html#p1029248

11. What is the value of \(xy\)?
(1) \(3^x*5^y=75\)
(2) \(3^{(x-1)(y-2)}=1\)

Solution: tough-and-tricky-exponents-and-roots-questions-125967-20.html#p1029249
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IanStewart
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  14 Jan 2012, 18:16
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Bunuel wrote:
2. If x, y, and z are positive integers and \(xyz=2,700\). Is \(\sqrt{x}\) and integer?
(1) \(y\) is an even perfect square and \(z\) is an odd perfect cube.
(2) \(\sqrt{z}\) is not an integer.

Note: a perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square. Similarly a perfect cube, is an integer that can be written as the cube of some other integer. For example 27=3^3, is a perfect cube.

Make prime factorization of 2,700 --> \(xyz=2^2*3^3*5^2\).

(1) \(y\) is an even perfect square and \(z\) is an odd perfect cube --> if y is either 2^2 or 2^2*5^2 and \(z=3^3\) =odd perfect square then \(x\) must be a perfect square which makes \(\sqrt{x}\) an integer: \(x=5^2\) or \(x=1\). But if \(z=1^3\) =odd perfect square then \(x\) could be \(3^3\) which makes \(\sqrt{x}\) not an integer. Not sufficient.

(2) \(\sqrt{z}\) is not an integer. Clearly insufficient.

(1)+(2) As from (1) \(\sqrt{z}\neq{integer}\) then \(z\neq{1}\), therefore it must be \(3^3\) (from 1) --> \(x\) is a perfect square which makes \(\sqrt{x}\) an integer: \(x=5^2\) or \(x=1\). Sufficient.

Answer: C.


In the highlighted part above, a couple of possibilities are missing. If y is an even perfect square, then we know y is divisible by 2^2. It may or may not be divisible by 3^2 and 5^2 (two choices each way, so four possibilities). So y could have any of four values:

2^2
(2^2)(3^2)
(2^2)(5^2)
(2^2)(3^2)(5^2)

Each of these values is possible when z = 1. While it doesn't affect the answer, it is also possible for x to be equal to 3 here, if y is equal to (2^2)(3^2)(5^2) and z is equal to 1, and x can also be (3)(5^2). I think you also meant to write 'odd perfect cube' instead of 'odd perfect square' (blue highlights).
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Bunuel
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  15 Jan 2012, 01:39
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IanStewart wrote:
Bunuel wrote:
2. If x, y, and z are positive integers and \(xyz=2,700\). Is \(\sqrt{x}\) and integer?
(1) \(y\) is an even perfect square and \(z\) is an odd perfect cube.
(2) \(\sqrt{z}\) is not an integer.

Note: a perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square. Similarly a perfect cube, is an integer that can be written as the cube of some other integer. For example 27=3^3, is a perfect cube.

Make prime factorization of 2,700 --> \(xyz=2^2*3^3*5^2\).

(1) \(y\) is an even perfect square and \(z\) is an odd perfect cube --> if y is either 2^2 or 2^2*5^2 and \(z=3^3\) =odd perfect square then \(x\) must be a perfect square which makes \(\sqrt{x}\) an integer: \(x=5^2\) or \(x=1\). But if \(z=1^3\) =odd perfect square then \(x\) could be \(3^3\) which makes \(\sqrt{x}\) not an integer. Not sufficient.

(2) \(\sqrt{z}\) is not an integer. Clearly insufficient.

(1)+(2) As from (1) \(\sqrt{z}\neq{integer}\) then \(z\neq{1}\), therefore it must be \(3^3\) (from 1) --> \(x\) is a perfect square which makes \(\sqrt{x}\) an integer: \(x=5^2\) or \(x=1\). Sufficient.

Answer: C.


In the highlighted part above, a couple of possibilities are missing. If y is an even perfect square, then we know y is divisible by 2^2. It may or may not be divisible by 3^2 and 5^2 (two choices each way, so four possibilities). So y could have any of four values:

2^2
(2^2)(3^2)
(2^2)(5^2)
(2^2)(3^2)(5^2)

Each of these values is possible when z = 1. While it doesn't affect the answer, it is also possible for x to be equal to 3 here, if y is equal to (2^2)(3^2)(5^2) and z is equal to 1, and x can also be (3)(5^2). I think you also meant to write 'odd perfect cube' instead of 'odd perfect square' (blue highlights).


Yes, I know Ian. For (1) I just discussed two possible scenarios to get an YES and NO answers to discard this statement, rather than listing all possibilities as you did. Thank you though for elaborating more on other cases.
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devinawilliam83
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  17 Mar 2012, 05:28
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Bunuel cant sqrt[x] be negative eg .. sqrt[4] = + or - 2.. in that case how is a suffucuent





SonyGmat wrote:
Bunuel wrote:
4. If \(xyz\neq{0}\) is \((x^{-4})*(\sqrt[3]{y})*(z^{-2})<0\)?
(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\)
(2) \(y^3>\frac{1}{z{^4}}\)


if we can find the sign of y we can answer the question.

(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}>0\) since it is\([\sqrt[4]{x^2}]=[\sqrt[]{x}]\)

Sufficient

(2) \(y^3>\frac{1}{z{^4}}>0\) since \(z^4>0\)

Sufficient

Therefore, D
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  17 Mar 2012, 05:33
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devinawilliam83 wrote:
Bunuel cant sqrt[x] be negative eg .. sqrt[4] = + or - 2.. in that case how is a suffucuent


1. GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers.

2. Any nonnegative real number has a unique non-negative square root called the principal square root and unless otherwise specified, the square root is generally taken to mean the principal square root.

When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.

That is, \(\sqrt{25}=5\), NOT +5 or -5. In contrast, the equation \(x^2=25\) has TWO solutions, \(\sqrt{25}=+5\) and \(-\sqrt{25}=-5\).

So, remember: even roots have only non-negative value on the GMAT.

Hope it's clear.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  13 Apr 2012, 18:14
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Bunuel wrote:
2. If x, y, and z are positive integers and \(xyz=2,700\). Is \(\sqrt{x}\) and integer?
(1) \(y\) is an even perfect square and \(z\) is an odd perfect cube.
(2) \(\sqrt{z}\) is not an integer.

Note: a perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square. Similarly a perfect cube, is an integer that can be written as the cube of some other integer. For example 27=3^3, is a perfect cube.

Make prime factorization of 2,700 --> \(xyz=2^2*3^3*5^2\).

(1) \(y\) is an even perfect square and \(z\) is an odd perfect cube --> if \(y\) is either \(2^2\) or \(2^2*5^2\) and \(z=3^3=odd \ perfect \ square\) then \(x\) must be a perfect square which makes \(\sqrt{x}\) an integer: \(x=5^2\) or \(x=1\). But if \(z=1^3=odd \ perfect \ cube\) then \(x\) could be \(3^3\) which makes \(\sqrt{x}\) not an integer. Not sufficient.

(2) \(\sqrt{z}\) is not an integer. Clearly insufficient.

(1)+(2) As from (1) \(\sqrt{z}\neq{integer}\) then \(z\neq{1}\), therefore it must be \(3^3\) (from 1) --> \(x\) is a perfect square which makes \(\sqrt{x}\) an integer: \(x=5^2\) or \(x=1\). Sufficient.

Answer: C.



Bunuel can you please elaborate from where we have these values. I missed the link . x=1 and z=1^3

Thanks
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  14 Apr 2012, 11:30
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carcass wrote:
Bunuel wrote:
2. If x, y, and z are positive integers and \(xyz=2,700\). Is \(\sqrt{x}\) and integer?
(1) \(y\) is an even perfect square and \(z\) is an odd perfect cube.
(2) \(\sqrt{z}\) is not an integer.

Note: a perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square. Similarly a perfect cube, is an integer that can be written as the cube of some other integer. For example 27=3^3, is a perfect cube.

Make prime factorization of 2,700 --> \(xyz=2^2*3^3*5^2\).

(1) \(y\) is an even perfect square and \(z\) is an odd perfect cube --> if \(y\) is either \(2^2\) or \(2^2*5^2\) and \(z=3^3=odd \ perfect \ square\) then \(x\) must be a perfect square which makes \(\sqrt{x}\) an integer: \(x=5^2\) or \(x=1\). But if \(z=1^3=odd \ perfect \ cube\) then \(x\) could be \(3^3\) which makes \(\sqrt{x}\) not an integer. Not sufficient.

(2) \(\sqrt{z}\) is not an integer. Clearly insufficient.

(1)+(2) As from (1) \(\sqrt{z}\neq{integer}\) then \(z\neq{1}\), therefore it must be \(3^3\) (from 1) --> \(x\) is a perfect square which makes \(\sqrt{x}\) an integer: \(x=5^2\) or \(x=1\). Sufficient.

Answer: C.



Bunuel can you please elaborate from where we have these values. I missed the link . x=1 and z=1^3

Thanks


Are you talking about statement 1? The values of x, y, and z analyzed there are just possible values that satisfy statement 1.

Given that: \(xyz=2^2*3^3*5^2\) and (1) says that \(y\) is an even perfect square and \(z\) is an odd perfect cube.

Now, if for example \(y=2^2*5^2=even \ perfect \ square\) and \(z=1^3=odd \ perfect \ cube\) then \(x=3^3\). You can apply the similar logic to obtain other possible values of x, y, and z.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  05 Sep 2013, 11:30
Bunuel wrote:
1. If \(357^x*117^y=a\), where \(x\) and \(y\) are positive integers, what is the units digit of \(a\)?
(1) \(100<y^2<x^2<169\)
(2) \(x^2-y^2=23\)

(1) \(100<y^2<x^2<169\) --> since both \(x\) and \(y\) are positive integers then \(x^2\) and \(y^2\) are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 --> \(y=11\) and \(x=12\). Sufficient.(As cyclicity of units digit of \(7\) in integer power is \(4\), therefore the units digit of \(7^{23}\) is the same as the units digit of \(7^3\), so 3).

(2) \(x^2-y^2=23\) --> \((x-y)(x+y)=23=prime\) --> since both \(x\) and \(y\) are positive integers then: \(x-y=1\) and \(x+y=23\) --> \(y=11\) and \(x=12\). Sufficient.

Answer: D.


Can anybody clarify for me how X-Y = 1 in the 2nd part of the question?
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  05 Sep 2013, 11:35
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Acropora wrote:
Bunuel wrote:
1. If \(357^x*117^y=a\), where \(x\) and \(y\) are positive integers, what is the units digit of \(a\)?
(1) \(100<y^2<x^2<169\)
(2) \(x^2-y^2=23\)

(1) \(100<y^2<x^2<169\) --> since both \(x\) and \(y\) are positive integers then \(x^2\) and \(y^2\) are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 --> \(y=11\) and \(x=12\). Sufficient.(As cyclicity of units digit of \(7\) in integer power is \(4\), therefore the units digit of \(7^{23}\) is the same as the units digit of \(7^3\), so 3).

(2) \(x^2-y^2=23\) --> \((x-y)(x+y)=23=prime\) --> since both \(x\) and \(y\) are positive integers then: \(x-y=1\) and \(x+y=23\) --> \(y=11\) and \(x=12\). Sufficient.

Answer: D.


Can anybody clarify for me how X-Y = 1 in the 2nd part of the question?


23 is a prime number, so it can be broken into the product of two positive multiples only in one way 23=1*23. Now, since x and y are positive integers, then x-y<x+y, thus x-y=1 and x+y=23.

Does this make sense?
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  28 Apr 2014, 12:10
Hi,

7. If x is a positive integer is \sqrt{x} an integer?
(1) \sqrt{7*x} is an integer
(2) \sqrt{9*x} is not an integer

when you explained the Stmt 1, I found it confusing.. However I took it in a different way. Let me know if my thinking is right.. For the 1st stmt to be right, x must be 7. So sqrt(7) is not a integer.. Sufficient..

I really missed the 2nd stmt when I did it.. Your way of solving and explaining is really good.. keep us posted with a more and more like these..
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  29 Apr 2014, 06:46
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sheolokesh wrote:
Hi,

7. If x is a positive integer is \sqrt{x} an integer?
(1) \(\sqrt{7*x}\) is an integer
(2) \(\sqrt{9*x}\)is not an integer

when you explained the Stmt 1, I found it confusing.. However I took it in a different way. Let me know if my thinking is right.. For the 1st stmt to be right, x must be 7. So sqrt(7) is not a integer.. Sufficient..

I really missed the 2nd stmt when I did it.. Your way of solving and explaining is really good.. keep us posted with a more and more like these..


For (1) x=7 is not the only possible value to satisfy \(\sqrt{7*x}=integer\). For example, x can also be 7*4, 7*9, 7*16, ... basically any number of a form 7*integer^2.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  01 Jul 2014, 09:20
Bunuel wrote:
1. If \(357^x*117^y=a\), where \(x\) and \(y\) are positive integers, what is the units digit of \(a\)?
(1) \(100<y^2<x^2<169\)
(2) \(x^2-y^2=23\)

(1) \(100<y^2<x^2<169\) --> since both \(x\) and \(y\) are positive integers then \(x^2\) and \(y^2\) are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 --> \(y=11\) and \(x=12\). Sufficient.(As cyclicity of units digit of \(7\) in integer power is \(4\), therefore the units digit of \(7^{23}\) is the same as the units digit of \(7^3\), so 3).

(2) \(x^2-y^2=23\) --> \((x-y)(x+y)=23=prime\) --> since both \(x\) and \(y\) are positive integers then: \(x-y=1\) and \(x+y=23\) --> \(y=11\) and \(x=12\). Sufficient.

Answer: D.


Hi Bunuel,

Could you please provide me some explanation for (2) when you say " since both x and y are positive integers then: x-y=1 and x+y=23 --> y=11 and x=12"?

How can we assume that x - y = 1? It could have been any two numbers. Is there a theory that if difference of square of two numbers, then they are consecutive? Could you please provide some reference?

Thanks for the help.

Regards,
Gajendra
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  01 Jul 2014, 09:33
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gkashyap wrote:
Bunuel wrote:
1. If \(357^x*117^y=a\), where \(x\) and \(y\) are positive integers, what is the units digit of \(a\)?
(1) \(100<y^2<x^2<169\)
(2) \(x^2-y^2=23\)

(1) \(100<y^2<x^2<169\) --> since both \(x\) and \(y\) are positive integers then \(x^2\) and \(y^2\) are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 --> \(y=11\) and \(x=12\). Sufficient.(As cyclicity of units digit of \(7\) in integer power is \(4\), therefore the units digit of \(7^{23}\) is the same as the units digit of \(7^3\), so 3).

(2) \(x^2-y^2=23\) --> \((x-y)(x+y)=23=prime\) --> since both \(x\) and \(y\) are positive integers then: \(x-y=1\) and \(x+y=23\) --> \(y=11\) and \(x=12\). Sufficient.

Answer: D.


Hi Bunuel,

Could you please provide me some explanation for (2) when you say " since both x and y are positive integers then: x-y=1 and x+y=23 --> y=11 and x=12"?

How can we assume that x - y = 1? It could have been any two numbers. Is there a theory that if difference of square of two numbers, then they are consecutive? Could you please provide some reference?

Thanks for the help.

Regards,
Gajendra


\((x-y)(x+y)=23=prime\). 23 is a prime number, so it can be broken into the product of two positive multiples only in one way 23=1*23. Now, since x and y are positive integers, then x-y<x+y, thus x-y=1 and x+y=23.

Does this make sense?
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  23 Jul 2014, 05:39
Bunuel wrote:
1. If \(357^x*117^y=a\), where \(x\) and \(y\) are positive integers, what is the units digit of \(a\)?
(1) \(100<y^2<x^2<169\)
(2) \(x^2-y^2=23\)

(1) \(100<y^2<x^2<169\) --> since both \(x\) and \(y\) are positive integers then \(x^2\) and \(y^2\) are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 --> \(y=11\) and \(x=12\). Sufficient.(As cyclicity of units digit of \(7\) in integer power is \(4\), therefore the units digit of \(7^{23}\) is the same as the units digit of \(7^3\), so 3).

(2) \(x^2-y^2=23\) --> \((x-y)(x+y)=23=prime\) --> since both \(x\) and \(y\) are positive integers then: \(x-y=1\) and \(x+y=23\) --> \(y=11\) and \(x=12\). Sufficient.

Answer: D.


Hello Bunuel, for the second statement, you seem to indicate that since the difference of the squares of two positive integers is a prime number, the two integers are consecutive on the number scale. Does this always hold?
Thank you!
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  23 Jul 2014, 08:29
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Kconfused wrote:
Bunuel wrote:
1. If \(357^x*117^y=a\), where \(x\) and \(y\) are positive integers, what is the units digit of \(a\)?
(1) \(100<y^2<x^2<169\)
(2) \(x^2-y^2=23\)

(1) \(100<y^2<x^2<169\) --> since both \(x\) and \(y\) are positive integers then \(x^2\) and \(y^2\) are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 --> \(y=11\) and \(x=12\). Sufficient.(As cyclicity of units digit of \(7\) in integer power is \(4\), therefore the units digit of \(7^{23}\) is the same as the units digit of \(7^3\), so 3).

(2) \(x^2-y^2=23\) --> \((x-y)(x+y)=23=prime\) --> since both \(x\) and \(y\) are positive integers then: \(x-y=1\) and \(x+y=23\) --> \(y=11\) and \(x=12\). Sufficient.

Answer: D.


Hello Bunuel, for the second statement, you seem to indicate that since the difference of the squares of two positive integers is a prime number, the two integers are consecutive on the number scale. Does this always hold?
Thank you!


Yes, positive integer solutions of x^2-y^2=prime must be consecutive, because of x-y=1 (x=y+1).
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  21 Aug 2014, 02:30
Bunuel wrote:
8. What is the value of \(x^2+y^3\)?
(1) \(x^6+y^9=0\)
(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\)

(1) \(x^6+y^9=0\) --> \((x^2)^3=(-y3)^3\) ---> \(x^2=-y^3\) --> \(x^2+y^3=0\). Sufficient.

(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\) --> \(3^{3x^2}=\frac{3}{3^{3y^2}*3}\) --> \(3^{3x^2}*3^{3y^2}=1\) --> \(3^{3x^2+3y^2}=1\) --> \(3x^2+3y^2=0\) (the power of 3 must be zero in order this equation to hold true) --> \(x^2+y^2=0\) the sum of two non-negative values is zero --> both \(x\) and \(y\) must be zero --> \(x=y=0\) --> \(x^2+y^3=0\). Sufficient.

Answer: D.
Hi Bunuel, I have a minor doubt. Using the corollary a^3+b^3=(a+b)(a^2-a*b+b^2) and plugging in a=x^2 and b=y^3 in statement 1,
we can say that (x^2)^3+(y^3)^3=(x^2+y^3)(x^4-x^2y^3+y^6)=0
Since any of expressions (x^2+y^3) or (x^4-x^2y^3+y^6) can be zero. It seems that statement 1 is insufficient by this logic.
Kindly comment.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  21 Aug 2014, 03:14
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joshnsit wrote:
Bunuel wrote:
8. What is the value of \(x^2+y^3\)?
(1) \(x^6+y^9=0\)
(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\)

(1) \(x^6+y^9=0\) --> \((x^2)^3=(-y3)^3\) ---> \(x^2=-y^3\) --> \(x^2+y^3=0\). Sufficient.

(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\) --> \(3^{3x^2}=\frac{3}{3^{3y^2}*3}\) --> \(3^{3x^2}*3^{3y^2}=1\) --> \(3^{3x^2+3y^2}=1\) --> \(3x^2+3y^2=0\) (the power of 3 must be zero in order this equation to hold true) --> \(x^2+y^2=0\) the sum of two non-negative values is zero --> both \(x\) and \(y\) must be zero --> \(x=y=0\) --> \(x^2+y^3=0\). Sufficient.

Answer: D.
Hi Bunuel, I have a minor doubt. Using the corollary a^3+b^3=(a+b)(a^2-a*b+b^2) and plugging in a=x^2 and b=y^3 in statement 1,
we can say that (x^2)^3+(y^3)^3=(x^2+y^3)(x^4-x^2y^3+y^6)=0
Since any of expressions (x^2+y^3) or (x^4-x^2y^3+y^6) can be zero. It seems that statement 1 is insufficient by this logic.
Kindly comment.


The point is that only real roots of x^4 - x^2y^3 + y^6 = 0 is x = y = 0, for which x^2 + y^3 = 0.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  09 Jan 2015, 04:28
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Bunuel wrote:
8. What is the value of \(x^2+y^3\)?
(1) \(x^6+y^9=0\)
(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\)

(1) \(x^6+y^9=0\) --> \((x^2)^3=(-y3)^3\) ---> \(x^2=-y^3\) --> \(x^2+y^3=0\). Sufficient.

(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\) --> \(3^{3x^2}=\frac{3}{3^{3y^2}*3}\) --> \(3^{3x^2}*3^{3y^2}=1\) --> \(3^{3x^2+3y^2}=1\) --> \(3x^2+3y^2=0\) (the power of 3 must be zero in order this equation to hold true) --> \(x^2+y^2=0\) the sum of two non-negative values is zero --> both \(x\) and \(y\) must be zero --> \(x=y=0\) --> \(x^2+y^3=0\). Sufficient.

Answer: D.


Hi Bunuel,
I have a confusion. The problem statement with option no. 1 can be written as follows:
if a^3 + b^3 = 0 (1st option) then what is the value of a + b ? (where a = x^2 and b = y^3).
Now, we know that (a^3 + b^3) = (a+b) * (a^2 - a*b + b^2). (a^3 + b^3) has 2 factors.
And according to the explanation, since, (a^3 + b^3) =0 => a^3 = -(b^3) => a = -b => a + b = 0. So, whenever a^3 + b^3 is 0, a+b will also be 0 !!!!
Can these steps be true always? Please reply... i am really confused.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  09 Jan 2015, 04:39
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aniteshgmat1101 wrote:
Bunuel wrote:
8. What is the value of \(x^2+y^3\)?
(1) \(x^6+y^9=0\)
(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\)

(1) \(x^6+y^9=0\) --> \((x^2)^3=(-y3)^3\) ---> \(x^2=-y^3\) --> \(x^2+y^3=0\). Sufficient.

(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\) --> \(3^{3x^2}=\frac{3}{3^{3y^2}*3}\) --> \(3^{3x^2}*3^{3y^2}=1\) --> \(3^{3x^2+3y^2}=1\) --> \(3x^2+3y^2=0\) (the power of 3 must be zero in order this equation to hold true) --> \(x^2+y^2=0\) the sum of two non-negative values is zero --> both \(x\) and \(y\) must be zero --> \(x=y=0\) --> \(x^2+y^3=0\). Sufficient.

Answer: D.


Hi Bunuel,
I have a confusion. The problem statement with option no. 1 can be written as follows:
if a^3 + b^3 = 0 (1st option) then what is the value of a + b ? (where a = x^2 and b = y^3).
Now, we know that (a^3 + b^3) = (a+b) * (a^2 - a*b + b^2). (a^3 + b^3) has 2 factors.
And according to the explanation, since, (a^3 + b^3) =0 => a^3 = -(b^3) => a = -b => a + b = 0. So, whenever a^3 + b^3 is 0, a+b will also be 0 !!!!
Can these steps be true always? Please reply... i am really confused.


Yes, if a^3 + b^3 = 0, then a + b = 0.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  12 Jul 2015, 15:28
Bunuel wrote:
4. If \(xyz\neq{0}\) is \((x^{-4})*(\sqrt[3]{y})*(z^{-2})<0\)?
(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\)
(2) \(y^3>\frac{1}{z{^4}}\)

\(xyz\neq{0}\) means that neither of unknown is equal to zero. Next, \((x^{-4})*(\sqrt[3]{y})*(z^{-2})=\frac{\sqrt[3]{y}}{x^4*z^2}\), so the question becomes: is \(\frac{\sqrt[3]{y}}{x^4*z^2}<0\)? Since \(x^4\) and \(z^2\) are positive numbers then the question boils down whether \(\sqrt[3]{y}<0\), which is the same as whether \(y<0\) (recall that odd roots have the same sign as the base of the root, for example: \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\)).

(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\) --> as even root from positive number (\(x^2\) in our case) is positive then \(\sqrt[5]{y}>\sqrt[4]{x^2}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

(2) \(y^3>\frac{1}{z{^4}}\) --> the same here as \(\frac{1}{z{^4}}>0\) then \(y^3>\frac{1}{z{^4}}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

Answer: D.



Hello sir,
It was not mentioned in the question stem whether x, y and z are integers. In this question, if x, y and z were non-integers, wouldn't the procedure be different? Would you please explain?

Thanks! :D
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  12 Jul 2015, 15:59
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ranaazad wrote:
Bunuel wrote:
4. If \(xyz\neq{0}\) is \((x^{-4})*(\sqrt[3]{y})*(z^{-2})<0\)?
(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\)
(2) \(y^3>\frac{1}{z{^4}}\)

\(xyz\neq{0}\) means that neither of unknown is equal to zero. Next, \((x^{-4})*(\sqrt[3]{y})*(z^{-2})=\frac{\sqrt[3]{y}}{x^4*z^2}\), so the question becomes: is \(\frac{\sqrt[3]{y}}{x^4*z^2}<0\)? Since \(x^4\) and \(z^2\) are positive numbers then the question boils down whether \(\sqrt[3]{y}<0\), which is the same as whether \(y<0\) (recall that odd roots have the same sign as the base of the root, for example: \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\)).

(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\) --> as even root from positive number (\(x^2\) in our case) is positive then \(\sqrt[5]{y}>\sqrt[4]{x^2}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

(2) \(y^3>\frac{1}{z{^4}}\) --> the same here as \(\frac{1}{z{^4}}>0\) then \(y^3>\frac{1}{z{^4}}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

Answer: D.




Hello sir,
It was not mentioned in the question stem whether x, y and z are integers. In this question, if x, y and z were non-integers, wouldn't the procedure be different? Would you please explain?

Thanks! :D


This question has a trick in that it will be true for both integers and non-integers alike:

1. x^2 > 0 no matter x=integer or not. Similarly, \(\sqrt[4]{x^2}\) will be positive for x=integer or non-integer. Try it out. Let x = 1 , \(\sqrt[4]{1^2} = 1 > 0\) and if x= 0.5, \(\sqrt[4]{0.5^2} = 0.707 >0.\)

Thus it does not matter whether you take x = integer or a fraction.

Coming back to the question,

Per statement 1, as mentioned above, \(\sqrt[4]{1^2} = 1 > 0\) ---> \(\sqrt[5]{y} > 0\) -----> y >0. Thus the statement \((x^{-4})*(\sqrt[3]{y})*(z^{-2})<0\) will be false. Thus this statement is sufficient.

Per statement 2, z^4 > 0 whether z =1 or z= 0.5 ----> 1/(z^4) >0 ----> y^3 >0---> y>0 (satisfies for both, y =1 or y = 0.25). Thus the given statement \((x^{-4})*(\sqrt[3]{y})*(z^{-2})<0\) will be false. Thus this statement is sufficient.

As both the given statements are sufficient, the correct answer is D.

whenever you are in doubt, always test the given statements/conditions with different values (integers, fractions , etc). This way you will get a better picture.
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